Question

If \( \tan ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+\tan ^{-1} \frac{1}{8}=\frac{\pi}{8} \), then \( \frac{49}{25} x^{2}= \)

Answer 1

\(\tan^{-1} \frac 15 + \tan^{-1} \frac 18 + \frac 12 \sec^{-1}x = \frac \pi 8\)

⇒ \(\tan^{-1} \left(\cfrac{\frac 15 + \frac 18}{1 - \frac 15 \times \frac 18}\right) = \frac \pi 8 - \frac 12\sec^{-1}x= \frac 14 (\frac \pi2-2\sec^{-1}x)\)

⇒ \(4\tan^{-1} \left(\frac{13}{39}\right) = \frac \pi 2 - 2\sec^{-1}x\)

⇒ \(2\left(2\tan^{-1} (\frac 13)+ \sec^{-1}x\right) = \frac \pi 2\)

⇒ \(\sec^{-1}x + \tan^{-1} \left(\frac{\frac 23}{1- \frac 19}\right) = \frac \pi 4\)

⇒ \(\tan^{-1}\left(\frac 23 \times \frac 98\right) + \sec^{-1}x = \frac \pi 4\)

⇒ \(\tan^{-1}\left(\frac 34\right) + \tan^{-1}\left(\frac{\sqrt{x^2 -1}}1\right) = \frac \pi 4\)

⇒ \(\tan^{-1} \left(\frac{\frac 34 +\sqrt{x^2 -1}}{1-\frac 34\sqrt{x^2-1}}\right) = \frac \pi 4\)

⇒ \(\frac{3 + 4\sqrt{x^2 -1}}{4 - 3\sqrt{x^2 -1}} = \tan(\frac \pi 4) = 1\)

⇒ \(3+4\sqrt{x^2 -1}= 4 - 3\sqrt{x^2 -1}\)

⇒ \(7\sqrt{x^2 -1} = 4-3 =1\)

⇒ \(x^2 - 1 = (\frac17)^2 = \frac 1{49}\)

⇒ \(x^2 = 1 + \frac 1{49} = \frac{50}{49}\)

⇒ \(\frac{49x^2}{25} = \frac{49}{25} \times \frac{50}{49} = 2\)